(o) (1 + sinα – cosα)2 + (1 – sinα + cosα)2 = 4(1 – sinα . cosα)
Expanding the left side of the equation, we have:
(1 + sinα – cosα)² + (1 – sinα + cosα)²
= (1 + sinα – cosα)(1 + sinα – cosα) + (1 – sinα + cosα)(1 – sinα + cosα)
= 1 + 2sinα - 2cosα + sin²α - 2sinαcosα + cos²α + 1 - 2sinα + 2cosα + sin²α - 2sinαcosα + cos²α
= 2 + 2sin²α - 4sinαcosα + 2cos²α
Using the trigonometric identity sin²α + cos²α = 1, we can rewrite the expression as:
2 + 2(1 - cos²α) - 4sinαcosα + 2cos²α
= 2 + 2 - 2cos²α - 4sinαcosα + 2cos²α
= 4 - 2cos²α - 4sinαcosα
Now, let's simplify the right side of the equation:
4(1 – sinα . cosα)
= 4 - 4sinαcosα
From here, we can see that the left side and the right side of the equation are equal, as both simplify to 4 - 2cos²α - 4sinαcosα.
Therefore, the given equation is true for all values of α.
To solve this equation step-by-step, let's start by expanding and simplifying both sides:
Left side: (1 + sinα – cosα)² + (1 – sinα + cosα)²
= (1 + sinα – cosα)(1 + sinα – cosα) + (1 – sinα + cosα)(1 – sinα + cosα)
= (1 + sinα – cosα)² + (1 – sinα + cosα)² + 2(1 + sinα – cosα)(1 – sinα + cosα)
Right side: 4(1 – sinα . cosα)
= 4 - 4sinα . cosα
Now, let's simplify further:
(1 + sinα – cosα)² + (1 – sinα + cosα)² + 2(1 + sinα – cosα)(1 – sinα + cosα) = 4 - 4sinα . cosα
Expanding the products:
(1 + sinα – cosα)² = (1 + sinα – cosα)(1 + sinα – cosα)
= 1 + sinα + sinα + sin²α - cosα - cosα - sinα.cosα + cos²α
(1 – sinα + cosα)² = (1 – sinα + cosα)(1 – sinα + cosα)
= 1 – sinα – sinα + sin²α + cosα – cosα + sinα.cosα + cos²α
Simplifying these equations:
(1 + sinα – cosα)² = 1 + 2sinα + sin²α - 2cosα - 2sinα.cosα + cos²α
(1 – sinα + cosα)² = 1 – 2sinα + sin²α + 2cosα + 2sinα.cosα + cos²α
Now we can substitute these simplifications back into the original equation:
(1 + 2sinα + sin²α - 2cosα - 2sinα.cosα + cos²α) + (1 – 2sinα + sin²α + 2cosα + 2sinα.cosα + cos²α) + 2(1 + sinα – cosα)(1 – sinα + cosα) = 4 - 4sinα . cosα
Rearranging and simplifying:
2 + 2sin²α + 2cos²α = 4 - 4sinα . cosα
Simplifying further:
2(sin²α + cos²α) + 2 = 4 - 4sinα . cosα
Since sin²α + cos²α = 1, the equation becomes:
2 + 2 = 4 - 4sinα . cosα
Simplifying once more:
4 = 4 - 4sinα . cosα
Now, subtracting 4 from both sides:
0 = -4sinα . cosα
Dividing both sides by -4:
0 = sinα . cosα
Therefore, the equation is satisfied when sinα . cosα = 0.